What is the foci ((x-3}{16})^2-\frac{(y-1)^2)/9 =1 ?

The foci ((x-3}{16})^2-\frac{(y-1)^2)/9 =1 is (3+sqrt(265),1),(3-sqrt(265),1)

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foci ((x-3}{16})^2-\frac{(y-1)^2)/9 =1

foci

(\frac{x - 3}{16})^{2} - \frac{( y - 1 ) ^{2}}{9} = 1

(3 + 265, 1), (3 - 265, 1)

Solution steps

(h + c, k), (h - c, k)

Calculate Hyperbola properties

(\frac{x - 3}{16})^{2} - \frac{( y - 1 ) ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (3, 1), a = 16, b = 3

(3 + c, 1), (3 - c, 1)

Compute

c :

c = 1 6^{2} + 3^{2} : 265

(3 + 265, 1), (3 - 265, 1)

9 x^{2} - 4 y^{2} - 18 x - 16 y - 43 = 0

9 x^{2} - 25 y^{2} = 225

\frac{x ^{2}}{9} - \frac{y ^{2}}{64} = 1

x = x^{2} - \frac{y ^{2}}{9}

11 x^{2} - 7 y^{2} = 77

What is the foci ((x-3}{16})^2-\frac{(y-1)^2)/9 =1 ?
The foci ((x-3}{16})^2-\frac{(y-1)^2)/9 =1 is (3+sqrt(265),1),(3-sqrt(265),1)